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Klio2033 [76]
2 years ago
13

The surface areas of two similar solids are 312 ft2 and 1,198 ft2. The volume of the smaller solid is 239 ft3. What is the volum

e of the larger solid? 1,798 ft3
Mathematics
2 answers:
marin [14]2 years ago
5 0

Answer:

The volume of the larger solid is 1,798\ ft^{3}

Step-by-step explanation:

step 1

Find the scale factor

we know that

If two figures are similar , then the ratio of its surface areas is equal to the scale factor squared

so

Let

z-------> the scale factor

x-------------> surface area larger solid

y-------------> surface area smaller solid

z^{2} =\frac{x}{y}

substitute

z^{2} =\frac{1,198}{312}

z=\sqrt{\frac{1,198}{312}} ----> scale factor

step 2

Find the volume of the larger solid

we know that

If two figures are similar , then the ratio of its volumes is equal to the scale factor elevated to the cube

so

Let

z-------> the scale factor

x-------------> volume of the  larger solid

y-------------> volume of the smaller solid

z^{3}=\frac{x}{y}

we have

z=\sqrt{\frac{1,198}{312}}

y=239\ ft^{3}

substitute the values

(\sqrt{\frac{1,198}{312}}^{3})=\frac{x}{239}

x=239*(\sqrt{\frac{1,198}{312}}^{3})=1,798\ ft^{3}

Molodets [167]2 years ago
4 0

Answer:

1,798 ft3

Step-by-step explanation:

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(b) ∠M is half the difference between the measures of arcs BE and BF, so is ...

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Step-by-step explanation:

We don't know the distribution for the scores. But we know the following properties:

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For this case we can use the Chebysev theorem who states that "At least 1 -\frac{1}{k^2} of the values lies between \mu -k\sigma and \mu +k\sigma"

And we need the boundaries on which we expect at least 75% of the scores. If we use the Chebysev rule we have this:

0.75 = 1-\frac{1}{k^2}

If we solve for k we can do this:

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